General Relativity and Cosmology for Undergraduates J. Norbury

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GENERAL RELATIVITY &
COSMOLOGY
for Undergraduates
Professor John W. Norbury
Physics Department
University of Wisconsin-Milwaukee
P.O. Box 413
Milwaukee, WI 53201
1997
Contents
1 NEWTONIAN COSMOLOGY 5
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2 Equation of State . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.1 Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2.2 Radiation . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Velocity and Acceleration Equations . . . . . . . . . . . . . . 7
1.4 Cosmological Constant . . . . . . . . . . . . . . . . . . . . . . 9
1.4.1 Einstein Static Universe . . . . . . . . . . . . . . . . . 11
2 APPLICATIONS 13
2.1 Conservation laws . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Age of the Universe . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 In°ation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.4 Quantum Cosmology . . . . . . . . . . . . . . . . . . . . . . . 16
2.4.1 Derivation of the Schro˜dinger equation . . . . . . . . . 16
2.4.2 Wheeler-DeWitt equation . . . . . . . . . . . . . . . . 17
2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.7 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.8 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3 TENSORS 23
3.1 Contravariant and Covariant Vectors . . . . . . . . . . . . . . 23
3.2 Higher Rank Tensors . . . . . . . . . . . . . . . . . . . . . . . 26
3.3 Review of Cartesian Tensors . . . . . . . . . . . . . . . . . . . 27
3.4 Metric Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.4.1 Special Relativity . . . . . . . . . . . . . . . . . . . . . 30
3.5 Christofiel Symbols . . . . . . . . . . . . . . . . . . . . . . . . 31
1
2 CONTENTS
3.6 Christofiel Symbols and Metric Tensor . . . . . . . . . . . . . 36
3.7 Riemann Curvature Tensor . . . . . . . . . . . . . . . . . . . 38
3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.10 Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.11 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
4 ENERGY-MOMENTUM TENSOR 45
4.1 Euler-Lagrange and Hamilton’s Equations . . . . . . . . . . . 45
4.2 Classical Field Theory . . . . . . . . . . . . . . . . . . . . . . 47
4.2.1 Classical Klein-Gordon Field . . . . . . . . . . . . . . 48
4.3 Principle of Least Action . . . . . . . . . . . . . . . . . . . . 49
4.4 Energy-Momentum Tensor for Perfect Fluid . . . . . . . . . . 49
4.5 Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . 51
4.6 Interacting Scalar Field . . . . . . . . . . . . . . . . . . . . . 51
4.7 Cosmology with the Scalar Field . . . . . . . . . . . . . . . . 53
4.7.1 Alternative derivation . . . . . . . . . . . . . . . . . . 55
4.7.2 Limiting solutions . . . . . . . . . . . . . . . . . . . . 56
4.7.3 Exactly Solvable Model of In°ation . . . . . . . . . . . 59
4.7.4 Variable Cosmological Constant . . . . . . . . . . . . . 61
4.7.5 Cosmological constant and Scalar Fields . . . . . . . . 63
4.7.6 Clariflcation . . . . . . . . . . . . . . . . . . . . . . . . 64
4.7.7 Generic In°ation and Slow-Roll Approximation . . . . 65
4.7.8 Chaotic In°ation in Slow-Roll Approximation . . . . . 67
4.7.9 Density Fluctuations . . . . . . . . . . . . . . . . . . . 72
4.7.10 Equation of State for Variable Cosmological Constant 73
4.7.11 Quantization . . . . . . . . . . . . . . . . . . . . . . . 77
4.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
5 EINSTEIN FIELD EQUATIONS 83
5.1 Preview of Riemannian Geometry . . . . . . . . . . . . . . . . 84
5.1.1 Polar Coordinate . . . . . . . . . . . . . . . . . . . . . 84
5.1.2 Volumes and Change of Coordinates . . . . . . . . . . 85
5.1.3 Difierential Geometry . . . . . . . . . . . . . . . . . . 88
5.1.4 1-dimesional Curve . . . . . . . . . . . . . . . . . . . . 89
5.1.5 2-dimensional Surface . . . . . . . . . . . . . . . . . . 92
5.1.6 3-dimensional Hypersurface . . . . . . . . . . . . . . . 96
5.2 Friedmann-Robertson-Walker Metric . . . . . . . . . . . . . . 99
5.2.1 Christofiel Symbols . . . . . . . . . . . . . . . . . . . . 101
CONTENTS 3
5.2.2 Ricci Tensor . . . . . . . . . . . . . . . . . . . . . . . . 102
5.2.3 Riemann Scalar and Einstein Tensor . . . . . . . . . . 103
5.2.4 Energy-Momentum Tensor . . . . . . . . . . . . . . . 104
5.2.5 Friedmann Equations . . . . . . . . . . . . . . . . . . 104
5.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
6 Einstein Field Equations 107
7 Weak Field Limit 109
8 Lagrangian Methods 111
4 CONTENTS
Chapter 1
NEWTONIAN
COSMOLOGY
1.1 Introduction
Many of the modern ideas in cosmology can be explained without the need
to discuss General Relativity. The present chapter represents an attempt to
do this based entirely on Newtonian mechanics. The equations describing
the velocity (called the Friedmann equation) and acceleration of the universe
are derived from Newtonian mechanics and also the cosmological constant
is introduced within a Newtonian framework. The equations of state are
also derived in a very simple way. Applications such as conservation laws,
the age of the universe and the in°ation, radiation and matter dominated
epochs are discussed.
1.2 Equation of State
In what follows the equation of state for non-relativistic matter and radiation
will be needed. In particular an expression for the rate of change of density,
_‰, will be needed in terms of the density ‰ and pressure p. (The deflnition
_x · dxdt , where t is time, is being used.) The flrst law of thermodynamics is
dU + dW = dQ (1.1)
where U is the internal energy, W is the work and Q is the heat transfer.
Ignoring any heat transfer and writing dW = Fdr = pdV where F is the
5
6 CHAPTER 1. NEWTONIAN COSMOLOGY
force, r is the distance, p is the pressure and V is the volume, then
dU = ¡pdV: (1.2)
Assuming that ‰ is a relativistic energy density means that the energy is
expressed as
U = ‰V (1.3)
from which it follows that
_U = _‰V + ‰ _V = ¡p _V (1.4)
where the term on the far right hand side results from equation (1.2). Writing
V / r3 implies that _VV = 3 _rr . Thus
_‰ = ¡3(‰+ p) _r
r
(1.5)
1.2.1 Matter
Writing the density of matter as
‰ =
M
4
3…r
3
(1.6)
it follows that
_‰ · d‰
dr
_r = ¡3‰ _r
r
(1.7)
so that by comparing to equation (1.5), it follows that the equation of state
for matter is
p = 0: (1.8)
This is the same as obtained from the ideal gas law for zero temperature.
Recall that in this derivation we have not introduced any kinetic energy, so
we are talking about zero temperature.
1.2.2 Radiation
The equation of state for radiation can be derived by considering radiation
modes in a cavity based on analogy with a violin string [12]. For a standing
wave on a string flxed at both ends
L =
n‚
2
(1.9)
1.3. VELOCITY AND ACCELERATION EQUATIONS 7
where L is the length of the string, ‚ is the wavelength and n is a positive
integer (n = 1; 2; 3:::::). Radiation travels at the velocity of light, so that
c = f‚ = f
2L
n
(1.10)
where f is the frequency. Thus substituting f = n2Lc into Planck’s formula
U = „h! = hf , where h is Planck’s constant, gives
U =
nhc
2
1
L
/ V ¡1=3: (1.11)
Using equation (1.2) the pressure becomes
p · ¡dU
dV
=
1
3
U
V
: (1.12)
Using ‰ = U=V , the radiation equation of state is
p =
1
3
‰: (1.13)
It is customary to combine the equations of state into the form
p =
°
3
‰ (1.14)
where ° · 1 for radiation and ° · 0 for matter.These equations of state
are needed in order to discuss the radiation and matter dominated epochs
which occur in the evolution of the Universe.
1.3 Velocity and Acceleration Equations
The Friedmann equation, which specifles the speed of recession, is obtained
by writing the total energy E as the sum of kinetic plus potential energy
terms (and using M = 43…r
3‰ )
E = T + V =
1
2
m _r2 ¡GMm
r
=
1
2
mr2(H2 ¡ 8…G
3
‰) (1.15)
where the Hubble constant H · _rr , m is the mass of a test particle in the
potential energy fleld enclosed by a gas of dust of mass M , r is the distance
from the center of the dust to the test particle and G is Newton’s constant.
8 CHAPTER 1. NEWTONIAN COSMOLOGY
Recall that the escape velocity is just vescape =
q
2GM
r =
q
8…G
3 ‰r
2, so that
the above equation can also be written
_r2 = v2escape ¡ k013¡ 2 (1.16)
with k0 · ¡2Em . The constant k0 can either be negative, zero or positive
corresponding to the total energy E being positive, zero or negative. For
a particle in motion near the Earth this would correspond to the particle
escaping (unbound), orbiting (critical case) or returning (bound) to Earth
because the speed _r would be greater, equal to or smaller than the escape
speed vescape. Later this will be analagous to an open, °at or closed universe.
Equation (1.15) is re-arranged as
H2 =
8…G
3
‰+
2E
mr2
:13¡ 3 (1.17)
Deflning k · ¡ 2E
ms2
and writing the distance in terms of the scale factor R
and a constant length s as r(t) · R(t)s, it follows that _rr =
_R
R and
r˜
r =
R˜
R ,
giving the Friedmann equation
H2 · (
_R
R
)2 =
8…G
3
‰¡ k
R2
(1.18)
which specifles the speed of recession. The scale factor is introduced because
in General Relativity it is space itself which expands [19]. Even though this
equation is derived for matter, it is also true for radiation. (In fact it is also
true for vacuum, with ⁄ · 8…G‰vac, where ⁄ is the cosmological constant
and ‰vac is the vacuum energy density which just replaces the ordinary den-
sity. This is discussed later.) Exactly the same equation is obtained from
the general relativistic Einstein fleld equations [13]. According to Guth [10],
k can be rescaled so that instead of being negative, zero or positive it takes
on the values ¡1; 0 or +1. From a Newtonian point of view this corresponds
to unbound, critical or bound trajectories as mentioned above. From a geo-
metric, general relativistic point of view this corresponds to an open, °at or
closed universe.
In elementary mechanics the speed v of a ball dropped from a height r
is evaluated from the conservation of energy equation as v =
p
2gr, where
g is the acceleration due to gravity. The derivation shown above is exactly
analagous to such a calculation. Similarly the acceleration a of the ball is
calculated as a = g from Newton’s equation F = mr˜, where F is the force
1.4. COSMOLOGICAL CONSTANT 9
and the acceleration is r˜ · d2r
dt2
. The acceleration for the universe is obtained
from Newton’s equation
¡GMm
r2
= mr˜:13¡ 5 (1.19)
Again using M = 43…r
3‰ and r˜r =
R˜
R gives the acceleration equation
F
mr
· r˜
r
· R˜
R
= ¡4…G
3
‰: (1.20)
However because M = 43…r
3‰ was used, it is clear that this acceleration
equation holds only for matter. In our example of the falling ball instead of
the acceleration being obtained from Newton’s Law, it can also be obtained
by taking the time derivative of the energy equation to give a = dvdt = v
dv
dr =
(
p
2gr)(
p
2g 1
2
p
r
) = g. Similarly, for the general case one can take the time
derivative of equation (1.18) (valid for matter and radiation)
d
dt
_R2 = 2 _RR˜ =
8…G
3
d
dt
(‰R2): (1.21)
Upon using equation (1.5) the acceleration equation is obtained as
R˜
R
= ¡4…G
3
(‰+ 3p) = ¡4…G
3
(1 + °)‰ (1.22)
which reduces to equation (1.20) for the matter equation of state (° = 0).
Exactly the same equation is obtained from the Einstein fleld equations [13].
1.4 Cosmological Constant
In both Newtonian and relativistic cosmology the universe is unstable to
gravitational collapse. Both Newton and Einstein believed that the Universe
is static. In order to obtain this Einstein introduced a repulsive gravitational
force, called the cosmological constant, and Newton could have done exactly
the same thing, had he believed the universe to be flnite.
In order to obtain a possibly zero acceleration, a positive term (conven-
tionally taken as ⁄3 ) is added to the acceleration equation (1.22) as
R˜
R
= ¡4…G
3
(‰+ 3p) +
⁄
3
(1.23)
10 CHAPTER 1. NEWTONIAN COSMOLOGY
which, with the proper choice of ⁄ can give the required zero acceleration
for a static universe. Again exactly the same equation is obtained from the
Einstein fleld equations [13]. What has been done here is entirely equivalent
to just adding a repulsive gravitational force in Newton’s Law. The question
now is how this repulsive force enters the energy equation (1.18). Identifying
the force from
r˜
r
=
R˜
R
· Frepulsive
mr
· ⁄
3
(1.24)
and using
Frepulsive =
⁄
3
mr · ¡dV
dr
(1.25)
gives the potential energy as
Vrepulsive = ¡12
⁄
3
mr2 (1.26)
which is just a repulsive simple harmonic oscillator. Substituting this into
the conservation of energy equation
E = T + V =
1
2
m _r2 ¡GMm
r
¡ 1
2
⁄
3
mr2 =
1
2
mr2(H2 ¡ 8…G
3
‰¡ ⁄
3
) (1.27)
gives
H2 · (
_R
R
)2 =
8…G
3
‰¡ k
R2
+
⁄
3
: (1.28)
Equations (1.28) and (1.23) constitute the fundamental equations of motion
that are used in all discussions of Friedmann models of the Universe. Exactly
the same equations are obtained from the Einstein fleld equations [13].
Let us comment on the repulsive harmonic oscillator obtained above.
Recall one of the standard problems often assigned in mechanics courses.
The problem is to imagine that a hole has been drilled from one side of the
Earth, through the center and to the other side. One is to show that if a
ball is dropped into the hole, it will execute harmonic motion. The solution
is obtained by noting that whereas gravity is an inverse square law for point
masses M and m separated by a distance r as given by F = GMm
r2
, yet if one
of the masses is a continous mass distribution represented by a density then
F = G43…‰mr. The force rises linearly as the distance is increased because
the amount of matter enclosed keeps increasing. Thus the gravitational force
for a continuous mass distribution rises like Hooke’s law and thus oscillatory
solutions are encountered. This sheds light on our repulsive oscillator found
1.4. COSMOLOGICAL CONSTANT 11
above. In this case we want the gravity to be repulsive, but the cosmological
constant acts just like the uniform matter distribution.
Finally authors often write the cosmological constant in terms of a vac-
uum energy density as ⁄ · 8…G‰vac so that the velocity and acceleration
equations become
H2 · (
_R
R
)2 =
8…G
3
‰¡ k
R2
+
⁄
3
=
8…G
3
(‰+ ‰vac)¡ k
R2
(1.29)
and
R˜
R
= ¡4…G
3
(1 + °)‰+
⁄
3
= ¡4…G
3
(1 + °)‰+
8…G
3
‰vac: (1.30)
1.4.1 Einstein Static Universe
Although we have noted that the cosmological constant provides repulsion,
it is interesting to calculate its exact value for a static universe [14, 15]. The
Einstein static universe requires R = R0 = constant and thus _R = R˜ = 0.
The case R˜ = 0 will be examined flrst. From equation (1.23) this requires
that
⁄ = 4…G(‰+ 3p) = 4…G(1 + °)‰: (1.31)
If there is no cosmological constant (⁄ = 0) then either ‰ = 0 which is an
empty universe, or p = ¡13‰ which requires negative pressure. Both of these
alternatives were unacceptable to Einstein and therefore he concluded that
a cosmological constant waspresent, i.e. ⁄ 6= 0. From equation (1.31) this
implies
‰ =
⁄
4…G(1 + °)
(1.32)
and because ‰ is positive this requires a positive ⁄. Substituting equa-
tion (1.32) into equation (1.28) it follows that
⁄ =
3(1 + °)
3 + °
[(
_R
R0
)2 +
k
R20
]: (1.33)
Now imposing _R = 0 and assuming a matter equation of state (° = 0)
implies ⁄ = k
R20
. However the requirement that ⁄ be positive forces k = +1
giving
⁄ =
1
R20
= constant: (1.34)
12 CHAPTER 1. NEWTONIAN COSMOLOGY
Thus the cosmological constant is not any old value but rather simply the
inverse of the scale factor squared, where the scale factor has a flxed value
in this static model.
Chapter 2
APPLICATIONS
2.1 Conservation laws
Just as the Maxwell equations imply the conservation of charge, so too do
our velocity and acceleration equations imply conservation of energy. The
energy-momentum conservation equation is derived by setting the covariant
derivative of the energy momentum tensor equal to zero. The same result is
achieved by taking the time derivative of equation (1.29). The result is
_‰+ 3(‰+ p)
_R
R
= 0: (2.1)
This is identical to equation (1.5) illustrating the intersting connection be-
tweeen thermodynamics and General Relativity that has been discussed re-
cently [16]. The point is that we used thermodynamics to derive our velocity
and acceleration equations and it is no surprise that the thermodynamic for-
mula drops out again at the end. However, the velocity and acceleration
equations can be obtained directly from the Einstein fleld equations. Thus
the Einstein equations imply this thermodynamic relationship in the above
equation.
The above equation can also be written as
d
dt
(‰R3) + p
dR3
dt
= 0 (2.2)
and from equation (1.14), 3(‰+ p) = (3 + °)‰, it follows that
d
dt
(‰R3+°) = 0: (2.3)
13
14 CHAPTER 2. APPLICATIONS
Integrating this we obtain
‰ =
c
R3+°
(2.4)
where c is a constant. This shows that the density falls as 1
R3
for matter and
1
R4
for radiation as expected.
Later we shall use these equations in a difierent form as follows. From
equation (2.1),
‰0 + 3(‰+ p)
1
R
= 0 (2.5)
where primes denote derivatives with respect to R, i.e. x0 · dx=dR. Alter-
natively
d
dR
(‰R3) + 3pR2 = 0 (2.6)
so that
1
R3+°
d
dR
(‰R3+°) = 0 (2.7)
which is consistent with equation (2.4)
2.2 Age of the Universe
Recent measurements made with the Hubble space telescope [17] have de-
termined that the age of the universe is younger than globular clusters. A
possible resolution to this paradox involves the cosmological constant [18].
We illustrate this as follows.
Writing equation (1.28) as
_R2 =
8…G
3
(‰+ ‰vac)R2 ¡ k (2.8)
the present day value of k is
k =
8…G
3
(‰0 + ‰0vac)R20 ¡H20R20 (2.9)
with H2 · ( _RR)2. Present day values of quantities have been denoted with a
subscript 0. Substituting equation (2.9) into equation (2.8) yields
_R2 =
8…G
3
(‰R2 ¡ ‰0R20 + ‰vacR2 ¡ ‰0vacR20)¡H20R20: (2.10)
2.3. INFLATION 15
Integrating gives the expansion age
T0 =
Z R0
0
dR
_R
=
Z R0
0
dRq
8…G
3 (‰R
2 ¡ ‰0R20 + ‰vacR2 ¡ ‰0vacR20)¡H20R20
:
(2.11)
For the cosmological constant ‰vac = ‰0vac and because R2 < R20 then a
non zero cosmological constant will give an age larger than would have been
obtained were it not present. Our aim here is simply to show that the
inclusion of a cosmological constant gives an age which is larger than if no
constant were present.
2.3 In°ation
In this section only a °at k = 0 universe will be discussed. Results for
an open or closed universe can easily be obtained and are discussed in the
references [13].
Currently the universe is in a matter dominated phase whereby the dom-
inant contribution to the energy density is due to matter. However the early
universe was radiation dominated and the very early universe was vacuum
dominated. Setting k = 0, there will only be one term on the right hand
side of equation (1.29) depending on what is dominating the universe. For a
matter (° = 0) or radiation (° = 1) dominated universe the right hand side
will be of the form 1
R3+°
(ignoring vacuum energy), whereas for a vacuum
dominated universe the right hand side will be a constant. The solution
to the Friedmann equation for a radiation dominated universe will thus be
R / t 12 , while for the matter dominated case it will be R / t 23 . One can see
that these results give negative acceleration, corresponding to a decelerating
expanding universe.
In°ation [19] occurs when the vacuum energy contribution dominates the
ordinary density and curvature terms in equation (1.29). Assuming these
are negligible and substituting ⁄ = constant, results in R / exp(t). The
acceleration is positive, corresponding to an accelerating expanding universe
called an in°ationary universe.
16 CHAPTER 2. APPLICATIONS
2.4 Quantum Cosmology
2.4.1 Derivation of the Schro˜dinger equation
The Wheeler-DeWitt equation will be derived in analogy with the 1 dimen-
sional Schro˜dinger equation, which we derive herein for completeness. The
Lagrangian L for a single particle moving in a potential V is
L = T ¡ V (2.12)
where T = 12m _x
2 is the kinetic energy, V is the potential energy. The action
is S =
R
Ldt and varying the action according to –S = 0 results in the
Euler-Lagrange equation (equation of motion)
d
dt
(
@L
@ _x
)¡ @L
@x
= 0 (2.13)
or just
_P =
@L
@x
(2.14)
where
P · @L
@ _x
: (2.15)
(Note P is the momentum but p is the pressure.) The Hamiltonian H is
deflned as
H(P; x) · P _x¡ L( _x; x): (2.16)
For many situations of physical interest, such as a single particle moving in
a harmonic oscillator potential V = 12kx
2, the Hamiltonian becomes
H = T + V = P
2
2m
+ V = E (2.17)
where E is the total energy. Quantization is achieved by the operator re-
placements P ! P^ = ¡i @@x and E ! E^ = i @@t where we are leaving ofi
factors of „h and we are considering the 1-dimensional equation only. The
Schro˜dinger equation is obtained by writing the Hamiltonian as an operator
H^ acting on a wave function “ as in
H^“ = E^“ (2.18)
and making the above operator replacements to obtain
(¡ 1
2m
@2
@x2
+ V )“ = i
@
@t
“ (2.19)
which is the usual form of the 1-dimensional Schro˜dinger equation written
in conflguration space.
2.4. QUANTUM COSMOLOGY 17
2.4.2 Wheeler-DeWitt equation
The discussion of the Wheeler-DeWitt equation in the minisuperspace ap-
proximation [20, 21, 11, 22] is usually restricted to closed (k = +1) and
empty (‰ = 0) universes. Atkatz [11] presented a very nice discussion for
closed and empty universes. Herein we consider closed, open and °at and
non-empty universes. It is important to consider the possible presence of
matter and radiation as they might otherwise change the conclusions. Thus
presented below is a derivation of the Wheeler-DeWitt equation in the min-
isuperspace approximation which also includes matter and radiation and
arbitrary values of k.
The Lagrangian is
L = ¡•R3[(
_R
R
)2 ¡ k
R2
+
8…G
3
(‰+ ‰vac)] (2.20)
with • · 3…4G . The momentum conjugate to R is
P · @L
@ _R
= ¡•2R _R: (2.21)
Substituting L and P into the Euler-Lagrange equation, _P ¡ @L@R = 0, equa-
tion (1.29) is recovered. (Note the calculation of @L@R is simplifled by using
the conservation equation (2.5) with equation (1.14), namely ‰0 + ‰0vac =
¡(3 + °)‰=R). The Hamiltonian H · P _R¡ L is
H( _R;R) = ¡•R3[(
_R
R
)2 +
k
R2
¡ 8…G
3
(‰+ ‰vac)] · 0 (2.22)
which has been written in terms of _R to show explicitly that the Hamiltonian
is identically zero and is not equal to the total energy as before. (Compare
equation (1.29)). In terms of the conjugate momentum
H(P;R) = ¡•R3[ P
2
4•2R4+
k
R2
¡ 8…G
3
(‰+ ‰vac)] = 0 (2.23)
which, of course is also equal to zero. Making the replacement P ! ¡i @@R
and imposing H“ = 0 results in the Wheeler-DeWitt equation in the min-
isuperspace approximation for arbitrary k and with matter or radiation (‰
term) included gives
f¡ @
2
@R2
+
9…2
4G2
[(kR2 ¡ 8…G
3
(‰+ ‰vac)R4]g“ = 0: (2.24)
18 CHAPTER 2. APPLICATIONS
Using equation (2.4) the Wheeler-DeWitt equation becomes
f¡ @
2
@R2
+
9…2
4G2
[kR2 ¡ ⁄
3
R4 ¡ 8…G
3
cR1¡° ]g“ = 0: (2.25)
This just looks like the zero energy Schro˜dinger equation [21] with a potential
given by
V (R) = kR2 ¡ ⁄
3
R4 ¡ 8…G
3
cR1¡° : (2.26)
For the empty Universe case of no matter or radiation (c = 0) the po-
tential V (R) is plotted in Figure 1 for the cases k = +1; 0;¡1 respectively
corresponding to closed [21], open and °at universes. It can be seen that only
the closed universe case provides a potential barrier through which tunnel-
ing can occur. This provides a clear illustration of the idea that only closed
universes can arise through quantum tunneling [22]. If radiation (° = 1 and
c 6= 0) is included then only a negative constant will be added to the poten-
tial (because the term R1¡° will be constant for ° = 1) and these conclusions
about tunneling will not change. The shapes in Figure 1 will be identical
except that the whole graph will be shifted downwards by a constant with
the inclusion of radiation. (For matter (° = 0 and c 6= 0) a term growing
like R will be included in the potential which will only be important for very
small R and so the conclusions again will not be changed.) To summarize,
only closed universes can arise from quantum tunneling even if matter or
radiation are present.
2.5 Summary
2.6. PROBLEMS 19
2.6 Problems
2.1
20 CHAPTER 2. APPLICATIONS
2.7 Answers
2.1
2.8. SOLUTIONS 21
2.8 Solutions
2.1
2.2
22 CHAPTER 2. APPLICATIONS
Chapter 3
TENSORS
3.1 Contravariant and Covariant Vectors
Let us imagine that an ’ordinary’ 2-dimensional vector has components (x; y)
or (x1; x2) (read as x superscript 2 not x squared) in a certain coordinate
system and components (x; y) or (x1; x2) when that coordinate system is ro-
tated by angle µ (but with the vector remaining flxed). Then the components
are related by [1] ˆ
x
y
!
=
ˆ
cos µ sin µ
sin µ cos µ
!ˆ
x
y
!
(3.1)
Notice that we are using superscipts (xi) for the components of our or-
dinary vectors (instead of the usual subscripts used in freshman physics),
which henceforth we are going to name contravariant vectors. We empha-
size that these are just the ordinary vectors one comes across in freshman
physics.
Expanding the matrix equation we have
x = x cos µ + y sin µ (3.2)
y = ¡x sin µ + y cos µ
from which it follows that
@x
@x
= cos µ
@x
@y
= sin µ (3.3)
23
24 CHAPTER 3. TENSORS
@y
@x
= ¡ sin µ @y
@y
= cos µ
so that
x =
@x
@x
x+
@x
@y
y (3.4)
y =
@y
@x
x+
@y
@y
y
which can be written compactly as
xi =
@xi
@xj
xj (3.5)
where we will always be using the Einstein summation convention for doubly
repeated indices. (i.e. xiyi ·Pi xiyi)
Instead of deflning an ordinary (contravariant) vector as a little arrow
pointing in some direction, we shall instead deflne it as an object whose com-
ponents transform according to equation(3.5). This is just a fancy version
of equation(3.1), which is another way to deflne a vector as what happens
to the components upon rotation (instead of the deflnition of a vector as a
little arrow). Notice that we could have written down a diferential version
of (3.5) just from what we know about calculus. Using the inflnitessimal dxi
(instead of xi) it follows immediately that
dxi =
@xi
@xj
dxj (3.6)
which is identical to (3.5) and therefore we must say that dxi forms an
ordinary or contravariant vector (or an inflnitessimally tiny arrow).
While we are on the subject of calculus and inflnitessimals let’s think
about @
@xi
which is kind of like the ’inverse’ of dxi. From calculus if f =
f(x; y) and x = x(x; y) and y = y(x; y) (which is what (3.3) is saying) then
@f
@x
=
@f
@x
@x
@x
+
@f
@y
@y
@x
(3.7)
@f
@y
=
@f
@x
@x
@y
+
@f
@y
@y
@y
or simply
@f
@xi
=
@f
@xj
@xj
@xi
: (3.8)
3.1. CONTRAVARIANT AND COVARIANT VECTORS 25
Let’s ’remove’ f and just write
@
@xi
=
@xj
@xi
@
@xj
: (3.9)
which we see is similar to (3.5), and so we might expect that @=@xi are
the ’components’ of a ’non-ordinary’ vector. Notice that the index is in the
denominator, so instead of writing @=@xi let’s just always write it as xi for
shorthand. Or equivalently deflne
xi · @
@xi
(3.10)
Thus
xi =
@xj
@xi
xj : (3.11)
So now let’s deflne a contravariant vector A„ as anything whose components
transform as (compare (3.5))
A
„ · @x„@x”A”
(3.12)
and a covariant vector A„ (often also called a one-form, or dual vector or
covector)
A„ = @x
”
@x„A”
(3.13)
In calculus we have two fundamental objects dxi and the dual vector @=@xi.
If we try to form the dual dual vector @=@(@=@xi) we get back dxi [2]. A
set of points in a smooth space is called a manifold and where dxi forms a
space, @=@xi forms the corresponding ’dual’ space [2]. The dual of the dual
space is just the original space dxi. Contravariant and covariant vectors are
the dual of each other. Other examples of dual spaces are row and column
matrices (x y) and
ˆ
x
y
!
and the kets < aj and bras ja > used in quantum
mechanics [3].
Before proceeding let’s emphasize again that our deflnitions of contravari-
ant and covariant vectors in (3.13) and (3.13) are nothing more than fancy
versions of (3.1).
26 CHAPTER 3. TENSORS
3.2 Higher Rank Tensors
Notice that our vector components A„ have one index, whereas a scalar
(e.g. t = time or T = temperature) has zero indices. Thus scalars are called
tensors of rank zero and vectors are called tensors of rank one. We are
familiar with matrices which have two indices Aij . A contravariant tensor of
rank two is of the form A„” , rank three A„”° etc. A mixed tensor, e.g. A„” ,
is partly covariant and partly contravariant.
In order for an object to be called a tensor it must satisfy the tensor
transformation rules, examples of which are (3.13) and (3.13) and
T
„” = @x
„
@xfi
@x”
@xfl
Tfifl :
(3.14)
T
„
” =
@x„
@xfi
@xfl
@x”
Tfifl : (3.15)
T
„”
‰ =
@x„
@xfi
@x”
@xfl
@x°
@x‰
Tfifl° : (3.16)
Thus even though a matrix has two indices Aij , it may not necessarily be
a second rank tensor unless it satisfles the above tensor tranformation rules
as well. However all second rank tensors can be written as matrices.
Higher rank tensors can be constructed from lower rank ones by forming
what is called the outer product or tensor product [14] as follows. For instance
Tfifl · AfiBfl (3.17)
or
Tfifl°– · Afi°Bfl– : (3.18)
The tensor product is often written simply as
T = A›B (3.19)
(do Problem 3.1) (NNN Next time discuss wedge product - easy - just
introduce antisymmetry).
We can also construct lower rank tensors from higher rank ones by a
process called contraction, which sets a covariant and contravariant index
equal, and because of the Einstein summation convention equal or repeated
3.3. REVIEW OF CARTESIAN TENSORS 27
indices are summed over. Thus contraction represents setting two indices
equal and summing. For example
Tfifl°fl · Tfi° (3.20)
Thus contraction over a pair of indices reduces the rank of a tensor by two
[14].
The inner product [14] of two tensors is deflned by forming the outer
product and then contracting over a pair of indices as
Tfifl · Afi°B°fl : (3.21)
Clearlythe inner product of two vectors (rank one tensors) produces a scalar
(rank zero tensor) as
A„B„ = constant · A:B (3.22)
and it can be shown that A:B as deflned here is a scalar (do Problem 3.2).
A scalar is a tensor of rank zero with the very special transformation law of
invariance
c = c: (3.23)
It is easily shown, for example, that A„B„ is no good as a deflnition of inner
product for vectors because it is not invariant under transformations and
therefore is not a scalar.
3.3 Review of Cartesian Tensors
Let us review the scalar product that we used in freshman physics. We wrote
vectors as A = Aie^i and deflned the scalar product as
A:B · AB cos µ (3.24)
where A and B are the magnitudes of the vectors A and B and µ is the
angle between them. Thus
A:B = Aie^i:Bj e^j
= (e^i:e^j)AiBj
· gijAiBj (3.25)
28 CHAPTER 3. TENSORS
where the metric tensor gij is deflned as the dot product of the basis vectors.
A Cartesian basis is deflned as one in which gij · –ij (obtained from
e^i:e^j = je^ijje^j j cos µ = cos µ = –ij). That is, the basis vectors are of unit
length and perpendicular to each other in which case
A:B = AiBi
= AxBx +AyBy + :::: (3.26)
where the sum (+:::) extends to however many dimensions are being consid-
ered and
A:A · A2 = AiAi (3.27)
which is just Pythagoras’ theorem, A:A · A2 = AiAi = A2x +A2y + ::::::.
Notice that the usual results we learned about in freshman physics, equa-
tions (3.26) and (3.27), result entirely from requiring gij = –ij =
ˆ
1 0
0 1
!
in matrix notation.
We could easily have deflned a non-Cartesian space, for example, gij =ˆ
1 1
0 1
!
in which case Pythagoras’ theorem would change to
A:A · A2 = AiAi = A2x +A2y +AxAy: (3.28)
Thus it is the metric tensor gij · e^i:e^j given by the scalar product of the
unit vectors which (almost) completely deflnes the vector space that we are
considering. Now let’s return to vectors and one-forms (i.e. contravariant
and covariant vectors).
3.4 Metric Tensor
We have already seen (in Problem 3.2) that the inner product deflned by
A:B · A„B„ transforms as a scalar. (The choice A„B„ won’t do because
it is not a scalar). However based on the previous section, we would expect
that A:B can also be written in terms of a metric tensor. The most natural
way to do this is
A:B · A„B„
= g„”A”B„ (3.29)
assuming g„” is a tensor.
3.4. METRIC TENSOR 29
In fact deflning A:B · A„B„ · g„”A”B„ makes perfect sense because it
also transforms as a scalar (i.e. is invariant). (do Problem 3.3) Thus either
of the two right hand sides of (3.29) will do equally well as the deflnition of
the scalar product, and thus we deduce that
A„ = g„”A”
(3.30)
so that the metric tensor has the efiect of lowering indices. Similarly it can
raise indices
A„ = g„”A”
(3.31)
How is vector A written in terms of basis vectors ? Based on our expe-
rience with Cartesian vectors let’s deflne our basis vectors such that
A:B · A„B„
= g„”A”B„
· (e„:e”)A”B„ (3.32)
which imples that vectors can be written in terms of components and basis
vectors as
A = A„e„
= A„e„ (3.33)
Thus the basis vectors of a covariant vector (one-form) transform as con-
travariant vectors. Contravariant components have basis vectors that trans-
form as one-froms [5] (pg. 63-64).
The above results illuminate our °at (Cartesian) space results where
g„” · –„” , so that (3.31) becomes A„ = A„ showing that in °at space there
is no distinction between covariant and contravariant vectors. Because of
this it also follows that A = A„e^„ and A:B = A„B„ which were our °at
space results.
Two more points to note are the symmetry
g„” = g”„ (3.34)
30 CHAPTER 3. TENSORS
and the inverse deflned by
g„fig
fi” = –”„ = g
”
„ (3.35)
so that g”„ is the Kronecker delta. This follows by getting back what we start
with as in A„ = g„”A” = g„”g”fiAfi · –fi„Afi.
3.4.1 Special Relativity
Whereas the 3-dimensional Cartesian space is completely characterized by
g„” = –„” or
g„” =
0B@ 1 0 00 1 0
0 0 1
1CA (3.36)
Obviously for unit matrices there is no distinction between –”„ and –„” . The
4-dimensional spacetime of special relativity is specifled by
·„” =
0BBB@
1 0 0 0
0 ¡1 0 0
0 0 ¡1 0
0 0 0 ¡1
1CCCA (3.37)
If a contravariant vector is specifled by
A„ = (A0; Ai) = (A0;A) (3.38)
it follows that the covariant vector is A„ = ·„”A” or
A„ = (A0; Ai) = (A0;¡A) (3.39)
Note that A0 = A0.
Exercise: Prove equation (3.39) using (3.38) and (3.37).
Thus, for example, the energy momentum four vector p„ = (E;p) gives
p2 = E2 ¡ p2. Of course p2 is the invariant we identify as m2 so that
E2 = p2 +m2.
Because of equation (3.38) we must have
@„ · @
@x„
= (
@
@x0
;5) = ( @
@t
;5) (3.40)
implying that
@„ · @
@x„
= (
@
@x0
;¡5) = ( @
@t
;¡5) (3.41)
3.5. CHRISTOFFEL SYMBOLS 31
Note that @0 = @0 = @@t (with c · 1). We deflne
22 · @„@„ = @0@0 + @i@i = @0@0 ¡ @i@i
=
@2
@x02
¡52 = @
2
@t2
¡52 (3.42)
(Note that some authors [30] instead deflne 22 · 52 ¡ @2
@t2
).
Let us now brie°y discuss the fourvelocity u„ and proper time. We shall
write out c explicitly here.
Using dx„ · (cdt; dx) the invariant interval is
ds2 · dx„dx„ = c2dt2 ¡ dx2: (3.43)
The proper time ¿ is deflned via
ds · cd¿ = cdt
°
(3.44)
which is consistent with the time dilation efiect as the proper time is the
time measured in an observer’s rest frame. The fourvelocity is deflned as
u„ · dx
„
d¿
· (°c; °v) (3.45)
such that the fourmomentum is
p„ · mu„ = (E
c
;p) (3.46)
where m is the rest mass.
Exercise: Check that (mu„)2 = m2c2. (This must be true so that E2 =
(pc)2 + (mc2)2).
3.5 Christofiel Symbols
Some good references for this section are [7, 14, 8]. In electrodynamics in
°at spacetime we encounter
E = ¡~5` (3.47)
and
B = ~5£A (3.48)
32 CHAPTER 3. TENSORS
where E and B are the electric and magnetic flelds and ` and A are the scalar
and vectors potentials. ~5 is the gradient operator deflned (in 3 dimensions)
as
~5 · i^@=@x+ j^@=@y + k^@=@z
= e^1@=@x1 + e^2@=@x2 + e^3@=@x3: (3.49)
Clearly then ` and A are functions of x; y; z, i.e. ` = `(x; y; z) and
A = A(x; y; z). Therefore ` is called a scalar fleld and A is called a vector
fleld. E and B are also vector flelds because their values are a function of
position also. (The electric fleld of a point charge gets smaller when you
move away.) Because the left hand sides are vectors, (3.47) and (3.48) imply
that the derivatives ~5` and ~5£A also transform as vectors. What about
the derivative of tensors in our general curved spacetime ? Do they also
transform as tensors ?
Consider a vector fleld A„(x”) as a function of contravariant coordinates.
Let us introduce a shorthand for the derivative as
A„;” · @A„
@x”
(3.50)
We want to know whether the derivative A„;” is a tensor. That is does A„;”
transform according to A„;” = @x
fi
@x„
@xfl
@x”Afi;fl : ? To flnd out, let’s evaluate the
derivative explicitly
A„;” · @A„
@x”
=
@
@x”
(
@xfi
@x„
Afi)
=
@xfi
@x„
@Afi
@x”
+
@2xfi
@x”@x„
Afi (3.51)
but Afi is a function of x” not x” , i.e. Afi = Afi(x”) 6= Afi(x”) . Therefore
we must insert @Afi@x” =
@Afi
@x°
@x°
@x” so that
A„;” · @A„
@x”
=
@xfi
@x„
@x°
@x”
@Afi
@x°
+
@2xfi
@x”@x„
Afi
=
@xfi
@x„
@x°
@x”
Afi;° +
@2xfi
@x”@x„
Afi (3.52)
We see therefore that the tensor transformation law for A„;” is spoiled by
the second term. Thus A„;” is not a tensor [8, 7, 14].
3.5. CHRISTOFFEL SYMBOLS 33
To see why this problem occurs we should look at the deflnition of the
derivative [8],
A„;” · @A„
@x”
= lim
dx!0
A„(x+ dx)¡A„(x)
dx”
(3.53)
or more properly [7, 14] as limdx°!0
A„(x°+dx°)¡A„(x°)
dx” .
The problem however with (3.53) is thatthe numerator is not a vector
because A„(x+ dx) and A„(x) are located at difierent points. The difiernce
between two vectors is only a vector if they are located at the same point.
The difierence betweeen two vectors located at separate points is not a vector
because the transformations laws (3.12) and (3.13) depend on position. In
freshman physics when we represent two vectors A and B as little arrows,
the difierence A ¡ B is not even deflned (i.e. is not a vector) if A and B
are at difierent points. We flrst instruct the freshman student to slide one
of the vectors to the other one and only then we can visualize the difierence
between them. This sliding is achieved by moving one of the vectors parallel
to itself (called parallel transport), which is easy to do in °at space. Thus
to compare two vectors (i.e. compute A¡B) we must flrst put them at the
same spacetime point.
Thus in order to calculate A„(x+ dx)¡A„(x) we must flrst deflne what
is meant by parallel transport in a general curved space. When we parallel
transport a vector in °at space its components don’t change when we move
it around, but they do change in curved space. Imagine standing on the
curved surface of the Earth, say in Paris, holding a giant arrow (let’s call
this vector A) vertically upward. If you walk from Paris to Moscow and keep
the arrow pointed upward at all times (in other words transport the vector
parallel to itself), then an astronaut viewing the arrow from a stationary
position in space will notice that the arrow points in difierent directions in
Moscow compared to Paris, even though according to you, you have par-
allel transported the vector and it still points vertically upward from the
Earth. Thus the astronaut sees the arrow pointing in a difierent direction
and concludes that it is not the same vector. (It can’t be because it points
difierently; it’s orientation has changed.) Thus parallel transport produces a
difierent vector. Vector A has changed into a difierent vector C.
To flx this situation, the astronaut communicates with you by radio and
views your arrow through her spacecraft window. She makes a little mark
on her window to line up with your arrow in Paris. She then draws a whole
series of parallel lines on her window and as you walk from Paris to Moscow
she keeps instructing you to keep your arrow parallel to the lines on her
34 CHAPTER 3. TENSORS
window. When you get to Moscow, she is satisfled that you haven’t rotated
your arrow compared to the markings on her window. If a vector is parallel
transported from an ’absolute’ point of view (the astronaut’s window), then
it must still be the same vector A, except now moved to a difierent point
(Moscow).
Let’s denote –A„ as the change produced in vector A„(xfi) located at xfi
by an inflnitessimal parallel transport by a distance dxfi. We expect –A„ to
be directly proportional to dxfi.
–A„ / dxfi (3.54)
We also expect –A„ to be directly proportional to A„; the bigger our arrow,
the more noticeable its change will be. Thus
–A„ / A”dxfi (3.55)
The only sensible constant of proportionality will have to have covariant „
and fi indices and a contravariant ” index as
–A„ · ¡”„fiA”dxfi (3.56)
where ¡”„fi are called Christofiel symbols or coe–cients of a–ne connection
or simply connection coe–cients . As Narlikar [7] points out, whereas the
metric tensor tells us how to deflne distance betweeen neighboring points, the
connection coe–cients tell us how to deflne parallelism betweeen neighboring
points.
Equation (3.56) deflnes parallel transport. –A„ is the change produced
in vector A„ by an inflnitessimal transport by a distance dxfi to produce a
new vector C„ · A„ + –A„. To obtain parallel transport for a contravariant
vector B„ note that a scalar deflned as A„B„ cannot change under parallel
transport. Thus [8]
–(A„B„) = 0 (3.57)
from which it follows that (do Problem 3.4)
–A„ · ¡¡„”fiA”dxfi: (3.58)
We shall also assume [8] symmetry under exchange of lower indices,
¡fi„” = ¡
fi
”„: (3.59)
(We would have a truly crazy space if this wasn’t true [8]. Think about it !)
3.5. CHRISTOFFEL SYMBOLS 35
Continuing with our consideration of A„(xfi) parallel transported an in-
flnitessimal distance dxfi, the new vector C„ will be
C„ = A„ + –A„: (3.60)
whereas the old vector A„(xfi) at the new position xfi+dxfi will be A„(xfi+
dxfi) . The difierence betweeen them is
dA„ = A„(xfi + dxfi)¡ [A„(xfi) + –A„] (3.61)
which by construction is a vector. Thus we are led to a new deflnition of
derivative (which is a tensor [8])
A„;” · dA„
dx”
= lim
dx!0
A„(x+ dx)¡ [A„(x) + –A„]
dx”
(3.62)
Using (3.53) in (3.61) we have dA„ =
@A„
@x” dx
” ¡ –A„ = @A„@x” dx” ¡ ¡†„fiA†dxfi
and (3.62) becomes A„;” · dA„dx” = @A„@x” ¡ ¡†„”A† (because dx
fi
dx” = –
fi
” ) which
we shall henceforth write as
A„;” · A„;” ¡ ¡†„”A†
(3.63)
where A„;” · @A„@x” . The derivative A„;” is often called the covariant deriva-
tive (with the word covariant not meaning the same as before) and one can
easily verify that A„;” is a second rank tensor (which will be done later in
Problem 3.5). From (3.58)
A„;” · A„;” + ¡„”†A†
(3.64)
For tensors of higher rank the results are, for example, [14, 8]
A„”;‚ · A„”;‚ + ¡„‚†A†” + ¡”‚†A„†
36 CHAPTER 3. TENSORS
(3.65)
and
A„”;‚ · A„”;‚ ¡ ¡†„‚A†” ¡ ¡†”‚A„† (3.66)
and
A„”;‚ · A„”;‚ + ¡„‚†A†” ¡ ¡†”‚A„† (3.67)
and
A„”fifl;‚ · A„”fifl;‚ + ¡„‚†A†”fifl + ¡”‚†A„†fifl ¡ ¡†fi‚A„”†fl ¡ ¡†fl‚A„”fi† : (3.68)
3.6 Christofiel Symbols and Metric Tensor
We shall now derive an important formula which gives the Christofiel symbol
in terms of the metric tensor and its derivatives [8, 14, 7]. The formula is
¡fifl° =
1
2g
fi†(g†fl;° + g†°;fl ¡ gfl°;†):
(3.69)
Another result we wish to prove is that
¡†„† = (ln
p¡g);„ = 12 [ln(¡g)];„
(3.70)
where
g · determinantjg„” j: (3.71)
Note that g 6= jg„” j. Let us now prove these results.
Proof of Equation (3.69). The process of covariant difierentiation should
never change the length of a vector. To ensure this means that the covariant
derivative of the metric tensor should always be identically zero,
g„”;‚ · 0: (3.72)
Applying (3.66)
g„”;‚ · g„”;‚ ¡ ¡†„‚g†” ¡ ¡†”‚g„† · 0 (3.73)
3.6. CHRISTOFFEL SYMBOLS AND METRIC TENSOR 37
Thus
g„”;‚ = ¡†„‚g†” + ¡
†
”‚g„† (3.74)
and permuting the „”‚ indices cyclically gives
g‚„;” = ¡†‚”g†„ + ¡
†
„”g‚† (3.75)
and
g”‚;„ = ¡†”„g†‚ + ¡
†
‚„g”† (3.76)
Now add (3.75) and (3.76) and subtract (3.74) gives [8]
g‚„;” + g”‚;„ ¡ g„”;‚ = 2¡†„”g‚† (3.77)
because of the symmetries of (3.59) and (3.34). Multiplying (3.77) by g‚fi
and using (3.34) and (3.35) (to give g‚†g‚fi = g†‚g‚fi = –fi† ) yields
¡fi„” =
1
2
g‚fi(g‚„;” + g”‚;„ ¡ g„”;‚): (3.78)
which gives (3.69). (do Problems 3.5 and 3.6).
Proof of equation (3.70) [14] (Appendix II) Using gfi†g†fl;fi = gfi†gflfi;†
(obtained using the symmetry of the metric tensor and swapping the names
of indices) and contracting over fi”, equation (3.69) becomes (flrst and last
terms cancel)
¡fifl;fi =
1
2
gfi†(g†fl;fi + g†fi;fl ¡ gflfi;†):
=
1
2
gfi†g†fi;fl (3.79)
Deflning g as the determinant jg„” j and using (3.35) it follows that
@g
@g„”
= gg„” (3.80)
a result which can be easily checked. (do Problem 3.7) Thus (3.79) be-
comes
¡fifl;fi =
1
2g
@g
@g‚fi
@g‚fi
@xfl
=
1
2g
@g
@xfl
=
1
2
@ ln g
@xfl
(3.81)
which is (3.70), where in (3.70) we write ln(¡g) instead of ln g because g is
always negative.
38 CHAPTER 3. TENSORS
3.7 Riemann Curvature Tensor
The Riemann curvature tensor is one of the most important tensors in gen-
eral relativity. If it is zero then it means that the space is °at. If it is
non-zero then we have a curved space. This tensor is most easily derived
by considering the order of double difierentiationon tensors [28, 2, 9, 7, 8].
Firstly we write in general
A„;fifl ·
@2A„
@xfi@xfl
(3.82)
and also when we write A„;fifl we again mean second derivative. Many authors
instead write A„;fifl · A„;fi;fl or A„;fifl · A„;fi;fl . We shall use either notation.
In general it turns out that even though A„;fifl = A
„
;flfi, however in general
it is true that A„;fifl 6= A„;flfi. Let us examine this in more detail. Firstly
consider the second derivative of a scalar `. A scalar does not change under
parallel transport therefore `;„ = `;„. From (3.63) we have (`;„ is a tensor,
not a scalar)
`;„;” = `;„;” = `;„;” ¡ ¡†„”`;† (3.83)
but because ¡†„” = ¡
†
”„ it follows that `;„” = `;”„ meaning that the order
of difierentiation does not matter for a scalar. Consider now a vector. Let’s
difierentiate equation (3.64). Note that A„;” is a second rank tensor, so we
use (3.67) as follows
A„;”;‚ = A
„
;”;‚ + ¡
„
‚†A
†
;” ¡ ¡†”‚A„;†
=
@
@x‚
(A„;”) + ¡
„
‚†A
†
;” ¡ ¡†”‚A„;†
= A„;”;‚ + ¡
„
”†;‚A
† + ¡„”†A
†
;‚ + ¡
„
‚†A
†
;” ¡ ¡†”‚A„;† (3.84)
Now interchange the order of difierentiation (just swap the ” and ‚ indices)
A„;‚;” = A
„
;‚;” + ¡
„
‚†;”A
† + ¡„‚†A
†
;” + ¡
„
ӠA
†
;‚ ¡ ¡†‚”A„;† (3.85)
Subtracting we have
A„;”;‚ ¡A„;‚;” = A†(¡„”†;‚ ¡ ¡„‚†;” + ¡„‚µ¡µ”† ¡ ¡„”µ¡µ‚†)
· A†R„‚†” (3.86)
with the famous Riemann curvature tensor deflned as
3.8. SUMMARY 39
Rfifl°– · ¡¡fifl°;– + ¡fifl–;° + ¡fi†°¡†fl– ¡ ¡fi†–¡†fl°
(3.87)
Exercise: Check that equations (3.86) and (3.87) are consistent.
The Riemann tensor tells us everything essential about the curvature of
a space. For a Cartesian spcae the Riemann tensor is zero.
The Riemann tensor has the following useful symmetry properties [9]
Rfifl°– = ¡Rfifl–° (3.88)
Rfifl°– +R
fi
°–fl +R
fi
–fl° = 0 (3.89)
and
Rfifl°– = ¡Rflfi°– (3.90)
All other symmetry properties of the Riemann tensor may be obtained from
these. For example
Rfifl°– = R°–fifl (3.91)
Finally we introduce the Ricci tensor [9] by contracting on a pair of indices
Rfifl · R†fi†fl (3.92)
which has the property
Rfifl = Rflfi (3.93)
(It will turn out later that Rfifl = 0 for empty space [9] ). Note that the
contraction of the Riemann tensor is unique up to a sign, i.e. we could have
deflned R††fifl or R
†
fi†fl or R
†
fifl† as the Ricci tensor and we would have the
same result except that maybe a sign difiernce would appear. Thus difierent
books may have this sign difierence.
However all authors agree on the deflnition of the Riemann scalar (ob-
tained by contracting Rfifl)
R · Rfifi · gfiflRfifl (3.94)
Finally the Einstein tensor is deflned as
G„” · R„” ¡ 12Rg„” (3.95)
After discussing the stress-energy tensor in the next chapter, we shall put
all of this tensor machinery to use in our discussion of general relativity
following.
3.8 Summary
40 CHAPTER 3. TENSORS
3.9 Problems
3.1 If A„ and B” are tensors, show that the tensor product (outer product)
deflned by T„” · A„B” is also a tensor.
3.2 Show that the inner product A:B · A„B„ is invariant under transfor-
mations, i.e. show that it satisfles the tensor transformation law of a scalar
(thus it is often called the scalar product).
3.3 Show that the inner product deflned by A:B · g„”A„B” is also a scalar
(invariant under transformations), where g„” is assumed to be a tensor.
3.4 Prove equation (3.58).
3.5 Derive the transformation rule for ¡fifl° . Is ¡
fi
fl° a tensor ?
3.6 Show that A„;” is a second rank tensor.
3.7 Check that @g@g„” = gg
„” . (Equation (3.80)).
3.10. ANSWERS 41
3.10 Answers
no answers; only solutions
42 CHAPTER 3. TENSORS
3.11 Solutions
3.1
To prove that T„” is a tensor we must show that it satisfles the
tensor transformation law T„” =
@x„
@xfi
@xfl
@x” T
fi
fl .
Proof T„” = A
„
B” = @x
„
@xfiA
fi @xfl
@x”Bfl
= @x
„
@xfi
@xfl
@x”A
fiBfl
= @x
„
@xfi
@xfl
@x” T
fi
fl
QED.
3.2
First let’s recall that if f = f(µ; fi) and µ = µ(x; y) and fi =
fi(x; y) then @f@µ =
@f
@x
@x
@µ +
@f
@y
@y
@µ =
@f
@xi
@xi
@µ .
Now
A:B = A„B„ = @x
„
@xfiA
fi @xfl
@x„Bfl
= @x
„
@xfi
@xfl
@x„A
fiBfl
= @x
fl
@xfiA
fiBfl by the chain rule
= –flfiA
fiBfl
= AfiBfi
= A:B
3.11. SOLUTIONS 43
3.3
A:B · g„”A„B”
= @x
fi
@x„
@xfl
@x” gfifl
@x„
@x°
@x”
@x–
A°B–
= @x
fi
@x„
@xfl
@x”
@x„
@x°
@x”
@x–
gfiflA
°B–
= @x
fi
@x°
@xfl
@x–
gfiflA
°B–
= –fi° –
fl
– gfiflA
°B–
= gfiflAfiBfl
= AfiBfi
= A:B
3.4
3.5
3.6
3.7
44 CHAPTER 3. TENSORS
Chapter 4
ENERGY-MOMENTUM
TENSOR
It is important to emphasize that our discussion in this chapter is based
entirely on Special Relativity.
4.1 Euler-Lagrange and Hamilton’s Equations
Newton’s second law of motion is
F =
dp
dt
(4.1)
or in component form (for each component Fi)
Fi =
dpi
dt
(4.2)
where pi = m _qi (with qi being the generalized position coordinate) so that
dpi
dt = _m _qi + mq˜i. If _m = 0 then Fi = mq˜i = mai. For conservative forces
F = ¡5V where V is the scalar potential. Rewriting Newton’s law we have
¡dV
dqi
=
d
dt
(m _qi) (4.3)
Let us deflne the Lagrangian L(qi; _qi) · T ¡V where T is the kinetic energy.
In freshman physics T = T ( _qi) = 12m _q
2
i and V = V (qi) such as the harmonic
oscillator V (qi) = 12kq
2
i . That is in freshman physics T is a function only
of velocity _qi and V is a function only of position qi. Thus L(qi; _qi) =
45
46 CHAPTER 4. ENERGY-MOMENTUM TENSOR
T ( _qi) ¡ V (qi). It follows that @L@qi = ¡dVdqi and @L@ _qi = dTd _qi = m _qi = pi. Thus
Newton’s law is
Fi =
dpi
dt
@L
@qi
=
d
dt
(
@L
@ _qi
) (4.4)
with the canonical momentum [1] deflned as
pi · @L
@ _qi
(4.5)
The second equation of (4.4) is known as the Euler-Lagrange equations of
motion and serves as an alternative formulation of mechanics [1]. It is usually
written
d
dt
(
@L
@ _qi
)¡ @L
@qi
= 0 (4.6)
or just
_pi =
@L
@qi
(4.7)
We have obtained the Euler-Lagrange equations using simple arguments. A
more rigorous derivation is based on the calculus of variations [1] as discussed
in Section 7.3.
We now introduce the Hamiltonian H deflned as a function of p and q as
H(pi; qi) · pi _qi ¡ L(qi; _qi) (4.8)
For the simple case T = 12m _q
2
i and V 6= V ( _qi) we have pi @L@ _qi = m _qi so that
T = p
2
i
2m and pi _qi =
p2i
m so that H(pi; qi) =
p2i
2m + V (qi) = T + V which is the
total energy. Hamilton’s equations of motion immediately follow from (4.8)
as
@H
@pi
= _qi (4.9)
because L 6= L(pi) and @H@qi = ¡ @L@qi so that from (4.4)
¡@H
@qi
= _pi: (4.10)
4.2. CLASSICAL FIELD THEORY 47
4.2 Classical Field Theory
Scalar flelds are important in cosmology as they are thought to drive in°a-
tion. Such a fleld is called an in°aton, an example of which may be the Higgs
boson. Thus the fleld ` considered below can be thoguht of as an in°aton,
a Higgs boson or any other scalar boson.
In both special and general relativity we always seek covariant equations
in which space and time are given equal status. The Euler-Lagrange equa-
tions (4.6) are clearly not covariant because special emphasis is placed on
time via the _qi and ddt(
@L
@ _qi
) terms.
Let us replace the qi by a fleld ` · `(x) where x · (t;x). The generalized
coordiante q has been replaced by the fleld variable ` and the discrete index
i has been replaced by a continuously varying index x. In the next section
we shall show how to derive the Euler-Lagrange equations from the action
deflned as
S ·
Z
Ldt (4.11)
which again is clearly not covariant. A covariant form of the action would
involvea Lagrangian density L via
S ·
Z
Ld4x =
Z
Ld3xdt (4.12)
with L · R Ld3x. The term ¡ @L@qi in equation (4.6) gets replaced by the
covariant term ¡ @L@`(x) . Any time derivative ddt should be replaced with
@„ · @@x„ which contains space as well as time derivatives. Thus one can
guess that the covariant generalization of the point particle Euler-Lagrange
equations (4.6) is
@„
@L
@(@„`)
¡ @L
@`
= 0 (4.13)
which is the covariant Euler-Lagrange equation for scalar flelds. This will
be derived rigorously in the next section.
In analogy with the canonical momentum in equation (4.5) we deflne the
covariant momentum density
ƒ„ · @L
@(@„`)
(4.14)
so that the Euler-Lagrange equations become
@„ƒ„ =
@L
@`
(4.15)
48 CHAPTER 4. ENERGY-MOMENTUM TENSOR
The canonical momentum is deflned as
ƒ · ƒ0 = @L
@ _`
(4.16)
The energy momentum tensor is (analagous to (4.8))
T„” · ƒ„@”`¡ g„”L (4.17)
with the Hamiltonian density
H ·
Z
Hd3x
H · T00 = ƒ _` ¡ L (4.18)
4.2.1 Classical Klein-Gordon Field
In order to illustrate the foregoing theory we shall use the example of the
classical, massive Klein-Gordon fleld deflned with the Lagrangian density
(HL units ??)
LKG = 12(@„`@
„`¡m2`2)
=
1
2
[ _`2 ¡ (5`)2 ¡m2`2] (4.19)
The covariant momentum density is more easily evaluated by re-writing
LKG = 12g„”(@„`@”` ¡m2`2). Thus ƒ„ = @L@(@„`) = 12g„”(–fi„@”` + @„`–fi” )
= 12(–
fi
„@
„` + @”`–fi” ) =
1
2(@
fi` + @fi`) = @fi`. Thus for the Klein-Gordon
fleld we have
ƒfi = @fi` (4.20)
giving the canonical momentum ƒ = ƒ0 = @0` = @0` = _`,
ƒ = _` (4.21)
Evaluating @L@` = ¡m2`, the Euler-Lagrange equations give the fleld equation
as @„@„`+m2` or
(22 +m2)` = 0
˜`¡52`+m2` = 0 (4.22)
4.3. PRINCIPLE OF LEAST ACTION 49
which is the Klein-Gordon equation for a free, massive scalar fleld. In mo-
mentum space p2 = ¡22, thus
(p2 ¡m2)` = 0 (4.23)
(Note that some authors [30] deflne 22 · 52 ¡ @2
@t2
difierent from (3.42), so
that they write the Klein-Gordon equation as (22¡m2)` = 0 or (p2+m2)` =
0.)
The energy momentum tensor is
T„” · ƒ„@”`¡ g„”L
= @„`@”`¡ g„”L
= @„`@”`¡ 12g„”(@fi`@
fi`¡m2`2): (4.24)
Therefore the Hamiltonian density is H · T00 = _`2 ¡ 12(@fi`@fi` ¡ m2`2)
which becomes [31]
H = 1
2
_`2 +
1
2
(5`)2 + 1
2
m2`2
=
1
2
[ƒ2 + (5`)2 +m2`2] (4.25)
where we have relied upon the results of Section 3.4.1.
4.3 Principle of Least Action
derive EL eqns properly for q and ` (do later). Leave out for now.
4.4 Energy-Momentum Tensor for Perfect Fluid
The best references for this section are [9](Pg. 124-125), [7], and [32](Pg.
155). The book by D’Inverno [32] also has a nice discussion of the Navier-
Stokes equation and its relation to the material of this section. Other ref-
erences are [8](Pg. 83), [15](Pg. 330), [33](Pg. 259), [34](Pg. 38), and
[2].
These references show that the energy-momentum tensor for a perfect
°uid is
50 CHAPTER 4. ENERGY-MOMENTUM TENSOR
T„” = (‰+ p)u„u” ¡ p·„”
(4.26)
where ‰ is the energy density and p is the pressure. We shall now work this
out for several speciflc cases [9]. Fig. 2.5 of Narlikar’s book [7] is particularly
helpful.
Motionless dust represents a collection of particles at rest. Thus u„ =
(c;0), so that T 00 = ‰. The equation of state for dust is p = 0 so that
T ii = 0 = T 0i = T ij . Thus
T„” =
0BBB@
‰ 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
1CCCA (4.27)
Motionless °uid representes a collection of particles all moving randomly
(such that they exert a pressure) but the whole collection is at rest, such
as a gas of particles at non-zero temperature, but conflned in a motionless
container. In this case u„ = (c;0) again, but now p 6= 0. Thus again T 00 = ‰
but now T ii = p and T ij = 0 so that
T„” =
0BBB@
‰ 0 0 0
0 p 0 0
0 0 p 0
0 0 0 p
1CCCA (4.28)
Motionless radiation is characterized by the equation of state p = 13‰.
Again the radiation is conflned to a container not in motion so that u„ =
(°c;0). (The 13 just comes from randomizing the pressure in 3 dimensions
[7].) Thus
T„” =
4
3
‰u„u” ¡ 1
3
‰·„”
=
0BBB@
‰ 0 0 0
0 13‰ 0 0
0 0 13‰ 0
0 0 0 13‰
1CCCA (4.29)
Thus the general case is the motionless °uid energy-momentum tensor
in equation (4.28). The special cases of motionless dust or motionless radi-
ation are obtained with the respective substitutions of p = 0 or p = 13‰ in
equation (4.28).
4.5. CONTINUITY EQUATION 51
4.5 Continuity Equation
In classical electrodynamics the fourcurrent density is j„ · (c‰; j) and the
covariant conservation law is @„j„ = 0 which results in the equation of
continuity @‰@t + 5:j = 0. This can also be obtained from the Maxwell
equations by taking the divergence of Ampµere’s law. (do Problems 4.1
and 4.2) Thus the four Maxwell equations are entirely equivalent to only
three Maxwell equations plus the equation of continuity.
We had a similar situation in Chapters 1 and 2 where we found that the
velocity and acceleration equations imply the conservation equation. Thus
the two velocity and acceleration equations are entirely equivalent to only
the velocity equation plus the conservation law.
In analogy with electrodynamics the conservation law for the energy-
momentum tensor is
T„”;” = 0 (4.30)
In the next chapter we shall show how equation (2.1) can be derived from
this.
4.6 Interacting Scalar Field
We represent the interaction of a scalar fleld with a scalar potential V (`).
Recall our elementary results for L = T ¡ V = 12m _q2i ¡ V (qi) for the coordi-
nates _qi. These discrete coordinates _qi have now been replaced by continuous
fleld variables `(x) where ` has replaced the generalized coordinate q and
the discrete index i has been replaced by a continuous index x. Thus V (qi)
naturally gets replaced with V (`) where ` · `(x).
Thus for an interacting scalar fleld we simply tack on ¡V (`) to the free
Klein-Gordon Lagrangian of equation (4.19) to give
L = 1
2
(@„`@„`¡m2`2)¡ V (`)
· LO + LI (4.31)
where LO · LKG and LI · ¡V (`). Actually the Lagrangian of (4.31)
refers to a minimally coupled scalar fleld as opposed to conformally coupled
[21] (Pg. 276). It is important to emphasize that V (`) does not contain
derivative terms such as @„`. Thus the covariant momentum density and
canonical momentum remain the same as equations (4.20) and (4.21) for the
52 CHAPTER 4. ENERGY-MOMENTUM TENSOR
free particle case namely ƒfi = @fi` and … = _`. Solving the Euler-Lagrange
equations now gives
(22 +m2)`+ V 0 = 0
˜`¡52`+m2`+ V 0 = 0 (4.32)
with
V 0 · dV
d`
(4.33)
The energy-momentum tensor is the same as for the free particle case,
equation (4.24), except for the addition of g„”V (`) as in
T„” = @„`@”`¡ g„” [12(@fi`@
fi`¡m2`2)¡ V (`)] (4.34)
yielding the Hamiltonian density the same as for the free particle case, equa-
tion (4.25), except for the addition of V (`) as in
H · T00 = 12
_`2 +
1
2
(5`)2 + 1
2
m2`2 + V (`)
=
1
2
[ƒ2 + (5`)2 +m2`2] + V (`): (4.35)
The purely spatial components are Tii = @i`@i`¡gii[12(@fi`@fi`¡m2`2)¡
V (`)] and with gii = ¡1 (i.e. assume Special Relativity NNN) we obtain
Tii =
1
2
_`2 +
1
2
(5`)2 ¡ 1
2
m2`2 ¡ V (`) (4.36)
Note that even though Tii has repeated indices let us not assume
P
i is
implied in this case. That is Tii refers to Tii = T11 = T22 = T33 and not
Tii = T11 + T22 + T33. Some authors (e.g Serot and Walecka [34]) do assume
the latter convention and therefore will disagree with our results by 13 .
Let us assume that the efiects of the scalar fleld are averaged so as
to behave like a perfect (motionless) °uid. In that case, comparing equa-
tion (4.28), we make the identiflcation [13, 34]
E · ‰ ·< T00 > (4.37)
and
p ·< Tii > (4.38)
4.7. COSMOLOGY WITH THE SCALAR FIELD 53
where E · ‰ is the energydensity and p is the pressure. (Note that because
Serot and Walecka do assume the Einstein summation convention for Tii,
they actually write p · 13 < Tii >.) Making these identiflcations we have
‰ =
1
2
_`2 +
1
2
(5`)2 + 1
2
m2`2 + V (`) (4.39)
and
p =
1
2
_`2 +
1
2
(5`)2 ¡ 1
2
m2`2 ¡ V (`) (4.40)
Let us also assume that the scalar fleld is massless and that ` = `(t)
only, i.e. ` 6= `(x), so that spatial derivatives disappear. (See Pg. 276-277
of Kolb and Turner [21] and Pg. 138 of Islam [13]). Therefore we flnally
obtain [13, 21].
‰ = 12 _`
2 + V (`)
(4.41)
and
p = 12 _`
2 ¡ V (`)
(4.42)
4.7 Cosmology with the Scalar Field
We have flnished with our discussion of the energy-momentum tensor and
therefore we should now move onto the next chapter. However, with the
tools at hand (energy-momentum tensor and Friedmann equations) we can
discuss the relevance of the scalar fleld to cosmology without needing the
formalism of General Relativity. Therefore before proceeding to the next
chapter we shall make a brief digression and discuss the evolution of the
scalar fleld.
If one is considering cosmological evolution driven by a scalar fleld, one
can simply substitute the above expressions for ‰ and p into the Friedmann
54 CHAPTER 4. ENERGY-MOMENTUM TENSOR
and acceleration equations (1.29) and (1.30) to obtain the time evolution of
the scale factor as in
H2 · (
_R
R
)2 =
8…G
3
[
1
2
_`2 +
1
2
(5`)2 + 1
2
m2`2 + V (`)]¡ k
R2
+
⁄
3
(4.43)
and
R˜
R
= ¡4…G
3
[
1
2
_`2 +
1
2
(5`)2 ¡ 1
2
m2`2 ¡ V (`)] + ⁄
3
(4.44)
The equation for the time evolution of the scalar fleld is obtained either by
taking the time derivative of equation (4.43) or more simply by substituting
the expression for ‰ and p in equations (4.41) and (4.42) into the conservation
equation (2.1) to give
˜` + 3H[ _` +
(5`)2
_` ] +m
2`+ V 0 = 0: (4.45)
Note that this is a new Klein-Gordon equation quite difierent to equation (4.32).
The difierence occurs because we have now incorporated gravity via the
Friedmann and conservation equation. We shall derive this equation again
in Chapter 7.
Again assuming the fleld is massless and ignoring spatial derivatives we
have
˜` + 3H _` + V 0 = 0
(4.46)
Notice that this is the equation for a damped harmonic oscillator (V = 12kx
2
and dVdx · V 0 = kx with F = ¡V 0) as
mx˜+ d _x+ kx = 0 (4.47)
Kolb and Turner [21] actually also include a particle creation term due to
the decay of the scalar fleld, which will cause reheating, and instead write
˜` + 3H _` + ¡ _` + V 0 = 0 (4.48)
4.7. COSMOLOGY WITH THE SCALAR FIELD 55
4.7.1 Alternative derivation
We can derive the equation of motion (4.46) for the scalar fleld in a quicker
manner [29] (Pg. 73), but this derivation only seems to work if we set m = 0
and 5` = 0 at the beginning. (Exercise: flnd out what goes wrong if m 6= 0
and 5` 6= 0.)
Consider a Lagrangian for ` which already has the scale factor built into
it as
L = R3[1
2
(@„`@„`¡m2`2)¡ V (`)] (4.49)
The R3 factor comes from
p¡g = R3 for a Robertson-Walker metric. This
will be discussed in Chapter 7. Notice that it is the same factor which
sits outside the Friedmann Lagrangian in equation (2.20). The equation of
motion is (do Problem 4.3)
˜`¡52`+ 3H _` +m2`+ V 0 = 0 (4.50)
which is difierent to (4.45). (NNNN why ???) However if m = 0 and5` = 0
it is the same as (4.46).
Let’s only consider
L = R3[1
2
_`2 ¡ V (`)] (4.51)
which results from setting m = 0 and 5` = 0 in (4.49). The equation of
motion is
˜` + 3H _` + V 0 = 0 (4.52)
Notice how quickly we obtained this result rather than the long procedure
to get (4.46). We didn’t even use the energy-momentum tensor. Also realize
that because5` = 0 the above Lagrangian formalism is really no difierent to
our old fashioned formalism where we had qi(t). Here we have only ` = `(t)
(not ` = `(x)), and so we only have i = 1, i.e. qi · `.
Identifying the Lagrangian as [29] L = R3(T ¡ V ) we immediately write
down the total energy density ‰ = T + V = 12 _`
2 + V (`). Taking the time
derivative _‰ = _` ˜` + V 0 _` = ¡3H _`2 from (4.46) and substituting into the
conservation equation (2.1), _‰ = ¡3H(‰ + p) we obtain the pressure as
p = 12 _`
2 ¡ V (`). Thus our energy density and pressure derived here agree
with our results above (4.39) and (4.40). Notice that the pressure is nothing
more than p = L
R3
. [29].
56 CHAPTER 4. ENERGY-MOMENTUM TENSOR
4.7.2 Limiting solutions
Assuming that k = ⁄ = 0 the Friedmann equation becomes
H2 · (
_R
R
)2 =
8…G
3
(
1
2
_` + V ) (4.53)
This equation together with equation (4.46) form a set of coupled equations
where solutions give `(t) and R(t). We solve the coupled equations in the
standard way by flrst eliminating one variable, then solving one equation,
then substituting the solution back into the other equation to solve for the
other variable. Let’s write equation (4.46) purely in terms of ` by eliminating
R which appears in the form H = _RR . We eliminate R by substituting H
from (4.53) into (4.46) to give
˜` +
q
12…G( _`2 + 2V ) _`+V’=0
˜`2 + 2˜`V 0 ¡ 12…G( _`2 + 2V ) _`2 + V 02 = 0 (4.54)
Notice that this is a non-linear difierential equation for `, which is di–cult
to solve in general. In this section we shall study the solutions for certain
limiting cases. Once `(t) is obtained from (4.54) it is put back into (4.53)
to get R(t).
Potential Energy=0
Setting V = 0 we then have ‰ = 12 _`
2 = p. Thus our equation of state is
p = ‰ (4.55)
or ° = 3.
With V = V 0 = 0 we have
˜`2 +
p
12…G _`2 = 0 (4.56)
which has the solution (do problem 4.4)
`(t) = `o +
1p
12…G
ln[1 +
p
12…G _`(t¡ to)] (4.57)
4.7. COSMOLOGY WITH THE SCALAR FIELD 57
(Note that the solution is equation (9.18) of [29] is wrong.) Upon substitut-
ing this solution back into the Friedmann equation (4.53) and solving the
difierential equation we obtain (do problem 4.5)
R(t) = Ro[1 +
p
12…G _`o(t¡ to)]1=3: (4.58)
This result may be understood from another point of view. Writing the
Friedmann equations as
H2 · (
_R
R
)2 =
8…G
3
‰ (4.59)
and
‰ =
fi
Rm
(4.60)
then the solution is always
R/t2=m (4.61)
which always gives
‰ / 1
t2
: (4.62)
If
‰ = constant (4.63)
(corresponding to m = 0) then the solution is
R/et (4.64)
(do problem 4.6). Note that for m < 2, one obtains power law in°ation.
For ordinary matter (m = 3), or radiation (m = 4) we have R / t2=3 and
R / t1=2 respectively. Returning to the scalar fleld solution (4.57) the density
is ‰ = 12 _`
2 for V = 0. Thus
_`(t) =
_`
o
1 +
p
12…G _`o(t¡ to)
(4.65)
combined with ( RRo )
3 = 1 +
p
12…G _`o(t¡ to) from (4.58) yields
_`(t) =
_`
oR
3
o
R3
(4.66)
to give the density
58 CHAPTER 4. ENERGY-MOMENTUM TENSOR
‰ = 12
_`2
oR
6
o
R6
(4.67)
corresponding to m = 6 and thus R / t1=3 in agreement with (4.58). Note
also that this density ‰ / 1
R6
also gives ‰ / 1
t2
.
Thus for a scalar fleld with V = 0, we have p = ‰ (° = 3) and ‰ / 1
R6
.
Contrast this with matter for which p = 0(° = 0) and ‰ / 1
R3
or radiation
for which p = 13‰ (° = 1=3) and ‰ / 1R4 .
However equation (4.67) may not be interpreted as a decaying Cosmo-
logical Constant because p 6= ‰ (see later).
Kinetic Energy=0
Here we take _` = ˜` = 0, so that ‰ = V and p = ¡V giving
p = ¡‰ (4.68)
or ° = ¡3 which is a negative pressure equation of state. Our equation of
motion for the scalar fleld (4.54) becomes
V 0 = 0 (4.69)
meaning that
V = Vo (4.70)
which is constant. Substituting the solution into the Friedmann equation
(4.53) gives
H2 = (
_R
R
)2 =
8…G
3
Vo (4.71)
which acts as a Cosmological Constant and which has the solution(do prob-
lem 4.7)
R(t) = Roe
p
8…G
3
Vo(t¡to) (4.72)
which is an in°ationary solution, valid for any V .
Warning
We have found that if k = ⁄ = 0 and if ‰ / 1Rm then R/t2 for any
value of m. All of this is correct. To check this we might substitute into the
Friedmann equation as
H2 = (
_R
R
)2 / 1
t2
(4.73)
4.7. COSMOLOGY WITH THE SCALAR FIELD 59
and say _RR / 1t giving
R 1
R
dR
dt dt /
R dt
t which yields lnR / lnt and thus
R / t2=m. The result R /t is wrong because we have left out an important
constant.
Actually if _RR =
c
t then lnR = c ln t = ln t
c giving R / tc instead of R/t.
Let’s keep our constants then. Write ‰ = d
2
Rm then R = (
md
2 )
2=mt2=m
and ‰ = d
2
(md
2
)2t2
= (2=m)
2
t2
. Substituting into the Friedmann equation gives
( _RR)
2 = (2=m)
2
t2
or _RR =
(2=m)
t with the above constant C =
2
m yielding
R /t2=m in agreement with the correct result above.
The lesson is be careful of constants when doing back-of-the-envelope
calculations.
4.7.3 Exactly Solvable Model of In°ation
Because (4.54) is a di–cult non-linear equation, exactly solvable models are
very rare. We shall examine the model of Barrow [35] which can be solved
exactly and leads to power law in°ation. The advantage of an exactly solv-
able model is that one can develop ones physical intuition better. Barrow’s
model [35] is brie°y introduced by Islam [13].
Any scalar fleld model is specifled by writing down the potential V (`).
Barrow’s potential is
V (`) · fle¡‚` (4.74)
where fl and ‚ are constants to be determined. Barrow [35] claims that a
particular solution to (4.54) is (which was presumably guessed at, rather
then solving the difierential equation)
`(t) =
p
2Alnt (4.75)
where
p
2A is just some constant. We check this claim by substituting (4.74)
and (4.75) into (4.54). From this we flnd (do problem 4.9) that
‚ =
r
2
A
(4.76)
and
fl = ¡A (4.77)
or
fl = A(24…GA¡ 1) (4.78)
60 CHAPTER 4. ENERGY-MOMENTUM TENSOR
Note that Barrow is wrong when he writes ‚A =
p
2. Also he uses units
with 8…G = 1, so that the second solution (4.78), he writes correctly as
fl = A(3A¡ 1). Also Barrow doesn’t use the flrst solution (4.77) for reasons
we shall see shortly.
Having solved for `(t) we now substitute into (4.53) to solve for R(t).
(Recall `(t) and R(t) are the solutions we seek to our coupled equations
(4.46) and (4.53).) Substituting V = ⁄
t2
and _` =
p
2A
t (see solution to
problem 4.9) we have
H2 · (
_R
R
)2 =
8…G
3
(
1
2
2A
t2
+
fl
t2
) =
8…G
3
(A+ fl)
1
t2
(4.79)
giving an equivalent density
‰ =
A+ fl
t2
(4.80)
Clearly we see why we reject the flrst solution (4.77) with fl = ¡A. It would
give zero density. Using the second solution (4.78) with fl = A(24…GA¡ 1)
yields
‰ =
24…GA2
t2
: (4.81)
Solving the Friedmann equation (4.79) gives
R/t8…GA (4.82)
where D is some constant. Setting 8…G · 1 we have
R/tA (4.83)
in agreement with Barrow’s solution. Power law in°ation results for
A > 1: (4.84)
Inverting the solution (4.83) we have t2 = C 0R2=A where C 0 is some constant.
Substituting into (4.81) we have
‰ / 1
R2=A
(4.85)
which corresponds to a Weak decaying Cosmological Constant. (See sections
4.7.4 and 4.7.5) For the in°ationary result A > 1 we have 2A · m < 2 which
corresponds to the quantum tunneling solution!!
Note of course that (4.85) can also be obtained via ‰ = 13 _`
2+V. We have
V= fl
t2
/ 1
R2=A
and _` =
p
2A
t giving _`
2 / 1
R2=A
.
4.7. COSMOLOGY WITH THE SCALAR FIELD 61
4.7.4 Variable Cosmological Constant
In this section we address the question as to when the density can be inter-
preted as a Cosmological Constant. Recall the Friedmann equations
H2 · (
_R
R
)2 =
8…G
3
‰¡ k
R2
+
⁄
3
(4.86)
and
R˜
R
= ¡4…G
3
(‰+ 3p) +
⁄
3
= ¡4…G
3
‰(1 + °) +
⁄
3
(4.87)
for p = °3‰. Suppose ‰ = k = 0, then we have
H2 = (
_R
R
)2 =
⁄
3
(4.88)
and
R˜
R
=
⁄
3
(4.89)
where two things have happened. Firstly the velocity and acceleration equa-
tions both have the same right hand side. Secondly the acceleration is posi-
tive. What sort of density would give the same result. Again for k = 0
H2 = (
_R
R
)2 =
8…G
3
‰ (4.90)
and
R˜
R
=
8…G
3
‰ (4.91)
only for ° = ¡3 or p = ¡‰. From our conservation equation, _‰ = ¡3H(‰+p),
this can only happen for ‰=constant. Thus constant density with equation
of state p = ¡‰ acts identically to a Cosmological Constant. In addition
the solution is automatically are of exponential in°ation, R / eHt. (Ex-
ercise:verify this.) Let us deflne a Strong Cosmological Constant as one in
which the velocity and acceleration equations both have the same right hand
side, (which automatically implies that the acceleration is positive). Such a
Strong Cosmological Constant must be a true constant.
62 CHAPTER 4. ENERGY-MOMENTUM TENSOR
On the other hand we can imagine densities that still give a positive
acceleration (i.e. in°ation) but do not normally give the velocity and accel-
eration with the same right hand side. Examining (4.54) indicates that the
acceleration is guarenteed to be positive if ° < ¡1 giving p < ¡13‰. (Recall
that the exponential in°ation above required ° = ¡3, which is consistent
with the inequality.) Thus negative pressure gives in°ation. (Although not
all negative pressure gives in°ation, e.g. p = ¡14‰.) The in°ation due to
° < ¡1 will not be exponential in°ation, but something weaker like per-
haps power law in°ation. Let us deflne a Weak Cosmological Constant as
one which arises from negative pressure (actually p < ¡13‰) to give a positive
acceleration (in°ation) only. The velocity and acceleration equations need
not have the same right hand side.
Recall that ordinary matter and radiation, or any positive pressure equa-
tion of state, neccesarily leads to negative acceleration (with ⁄ = 0). Thus
positive pressure leads to attractive gravity. However positive acceleration
implies a repulsive gravity or antigravity. Thus negative pressure (actually
p < ¡13‰) leads to antigravity. This is why we wish to use the term weak Cos-
mological Constant (even though right hand sides are not the same) because
it is consonant with antigravity.
Let us summarize. We consider ‰ alone with ⁄ · 0. For
R˜ > 0) ° < ¡1 and ‰ =anything (e.g. ‰ = ‰(R) or ‰ =constant)
R˜ > 0 and ‰ =constant) ° = ¡3
For ° < ¡1, ‰ behaves as a weak Cosmological Constant and for ° = ¡3,
‰ behaves as a Strong Cosmological Constant.
Finally let us emphasize that it is perfectly legitimate to consider a Weak
Cosmological Constant as a real Cosmological Constant. Einstein’s original
motivation in introducing ⁄ was to obtain a static universe. Thus all he
wanted was a antigravity term; i.e. all he wanted was a weak Cosmological
Constant. It \accidentally" happened that the right hand sides turned out
to be equal, giving a strong ⁄.
Interpreting ‰ as a Cosmological Constant leads us to expect that a
weak Cosmological Constant can vary. This follows from ‰=anything above.
i.e. ‰=constant or ‰ = ‰(R) giving a variable function. (But a strong
Cosmological Constant cannot vary).
4.7. COSMOLOGY WITH THE SCALAR FIELD 63
4.7.5 Cosmological constant and Scalar Fields
Refer back to the density and pressure of the scalar fleld in equations (4.41))
and (4.42). We had
‰ =
1
2
_`2 + V (`)
and
‰ =
1
2
_`2 ¡ V (`):
For the case where PE = V = 0 we have p = ‰. The pressure is positive
and therefore equation (4.67), ‰ / 1
R6
, cannot be interpreted as a (variable)
Cosmological Constant.
For the case where KE = 12 _`
2 = 0 we have p = ¡‰ meaning that ‰ can
be interpreted as a Strong Cosmological Constant. (We